login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A206424 The number of 1's in row n of Pascal's Triangle (mod 3) 9
1, 2, 2, 2, 4, 4, 2, 4, 5, 2, 4, 4, 4, 8, 8, 4, 8, 10, 2, 4, 5, 4, 8, 10, 5, 10, 14, 2, 4, 4, 4, 8, 8, 4, 8, 10, 4, 8, 8, 8, 16, 16, 8, 16, 20, 4, 8, 10, 8, 16, 20, 10, 20, 28, 2, 4, 5, 4, 8, 10, 5, 10, 14, 4, 8, 10, 8, 16, 20, 10, 20, 28, 5, 10, 14, 10, 20, 28 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

A006047(n) = a(n) + A227428(n).

a(n) = n + 1 - A062296(n) - A227428(n); number of ones in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

LINKS

Reinhard Zumkeller (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..19683

R. Garfield, H. S. Wilf, The distribution of the binomial coefficients modulo p, J. Numb. Theory 41 (1) (1992) 1-5

Marcus Jaiclin, et al. Pascal's Triangle, Mod 2,3,5

D. L. Wells, Residue counts modulo three for the fibonacci triangle, Appl. Fib. Numbers, Proc. 6th Int Conf Fib. Numbers, Pullman, 1994 (1996) 521-536

A. Wilson, Pascal's Triangle Modulo 3, (2015)

FORMULA

From Antti Karttunen, Jul 27 2017: (Start)

a(n) = (3^k + 1)*2^(y-1), where y = A062756(n) and k = A081603(n). [See e.g. Wells or Wilson references.]

a(n) = A006047(n) - A227428(n).

(End)

From David A. Corneth and Antti Karttunen, Jul 27 2017: (Start)

Based on the first formula above, we have following identities:

a(3n) = a(n).

a(3n+1) = 2*a(n).

a(9n+4) = 4*a(n).

(End)

EXAMPLE

Example: Rows 0-8 of Pascal's Triangle (mod 3) are:

1                   So a(0) = 1

1 1                 So a(1) = 2

1 2 1               So a(2) = 2

1 0 0 1                 .

1 1 0 1 1               .

1 2 1 1 2 1             .

1 0 0 2 0 0 1

1 1 0 2 2 0 1 1

1 2 1 2 1 2 1 2 1

MATHEMATICA

Table[Count[Mod[Binomial[n, Range[0, n]], 3], 1], {n, 0, 99}] (* Alonso del Arte, Feb 07 2012 *)

PROG

(Haskell)

a206424 = length . filter (== 1) . a083093_row

-- Reinhard Zumkeller, Jul 11 2013

(PARI) A206424(n) = sum(k=0, n, 1==(binomial(n, k)%3)); \\ (naive way) Antti Karttunen, Jul 26 2017

(Scheme) (define (A206424 n) (* (+ (A000244 (A081603 n)) 1) (A000079 (- (A062756 n) 1)))) ;; (fast way) Antti Karttunen, Jul 27 2017

CROSSREFS

Cf. A083093, A062296, A006047, A062756, A081603, A206427, A227428.

Sequence in context: A139560 A192095 A105080 * A117728 A172309 A130453

Adjacent sequences:  A206421 A206422 A206423 * A206425 A206426 A206427

KEYWORD

nonn,easy

AUTHOR

Marcus Jaiclin, Feb 07 2012

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified February 23 18:13 EST 2019. Contains 320437 sequences. (Running on oeis4.)